Proof. We have to show the following. Given a scheme $V$ locally of finite type over $S$, given $\xi \in F(V)$, and given a finite type point $v_0 \in V$ such that $\xi$ is versal at $v_0$, after replacing $V$ by an open neighbourhood of $v_0$ we have that $\xi$ is versal at every finite type point of $V$. Write $\xi = (Z, u', \hat x)$.

First case: $v_0 \not\in Z$. Then we can first replace $V$ by $V \setminus Z$. Hence we see that $\xi = (\emptyset , u', \emptyset )$ and the morphism $u' : V \to X'$ is versal at $v_0$. By More on Morphisms of Spaces, Lemma 75.20.1 this means that $u' : V \to X'$ is smooth at $v_0$. Since the set of a points where a morphism is smooth is open, we can after shrinking $V$ assume $u'$ is smooth. Then the same lemma tells us that $\xi$ is versal at every point as desired.

Second case: $v_0 \in Z$. Write $W = \mathop{\mathrm{colim}}\nolimits W_ n$ as in Formal Spaces, Lemma 86.20.11. By Lemma 97.27.15 we may assume $\hat x : V_{/Z} \to W$ is a smooth morphism of formal algebraic spaces. It follows immediately that $\xi = (Z, u', \hat x)$ is versal at all finite type points of $Z$. Let $V' \to V$, $\hat x'$, and $x'$ witness the compatibility between $u'$ and $\hat x$. We see that $\hat x' : V'_{/Z} \to X'_{/T'}$ is smooth as a base change of $\hat x$. Since $\hat x'$ is the completion of $x' : V' \to X'$ this implies that $x' : V' \to X'$ is smooth at all points of $(V' \to V)^{-1}(Z) = |x'|^{-1}(T') \subset |V'|$ by the already used More on Morphisms of Spaces, Lemma 75.20.1. Since the set of smooth points of a morphism is open, we see that the closed set of points $B \subset |V'|$ where $x'$ is not smooth does not meet $(V' \to V)^{-1}(Z)$. Since $V' \to V$ is proper and hence closed, we see that $(V' \to V)(B) \subset V$ is a closed subset not meeting $Z$. Hence after shrinking $V$ we may assume $B = \emptyset$, i.e., $x'$ is smooth. By the discussion in the previous paragraph this exactly means that $\xi$ is versal at all finite type points of $V$ not contained in $Z$ and the proof is complete. $\square$

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